Hi,

I came up with the following construction for the given problem.
It checks out with Sketchpad. See what you think.

Bractals

PROBLEM:

Given an arbitrary point and triangle. Construct a line through the
point that divides the triangle into two polygons of equal area.

NOTATION:

^ denotes set intersection.

Let X' denote the midpoint of the side of the triangle opposite the vertex X.

If X and Y are distinct points, let

L(XY) denote the line determined by points X and Y,
R(XY) denote the ray starting at point X and passing through
point Y,
S(XY) denote the line segment determined by points X and Y, and
P(Z,XY) denote the line through point Z and parallel to L(XY).

L(YX) = L(XY), R(YX) != R(XY), S(YX) = S(XY)
S(XY) = R(XY) ^ R(YX)

CONSTRUCTION SCHEMA:

CIR(XYZ,r) : For fixed points Y and Z and fixed length r -
construct point X such that Y lies on S(XZ) and |XY| = r.

CONSTRUCTION:

If ( the point lies on L(XX') for some triangle vertex X ) {
L(XX') is the desired line.
} else {
Let P be the given point and G the centroid of the triangle.
Label the triangle ABC such that R(GP) intersects S(CB').
Construct point R such that R = P(P,AB) ^ L(AC).
Construct point S such that S = P(B',BR) ^ L(AB).

If ( P lies on the triangle ) {
L(PS) is the desired line.
} else {
Construct point T as the midpoint of S(AS).
CIR(QRP,|PR|).
CIR(DAT,|PQ|).
Construct a semicircle with diameter S(DT) on the opposite side
of S(DT) from vertex C.
Construct a line through vertex A, perpendicular to S(DT),
intersecting semicircle DT at point E.

If ( P is outside the triangle ) {
CIR(FTA,|TE|).
L(PF) is the desired line.
} else {
Construct point H such that |EH| = |AT| and H lies on S(AT).
CIR(KTH,|AH|).
L(PK) is the desired line.
}
}
}